Download App

Maths Solve the Following Question.(4 Marks)

Trigonometry – I · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 18 questions.

Questions

Solve the Following Question.(4 Marks)

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Find the trigonometric functions of : -210°
Answer

Image
Angle of measure $\left(-210^{\circ}\right)$ : Let $m \angle X O A=-210^{\circ}$ Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the X-axis. $\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. $O P=1$ Ur $=1$ $\begin{aligned} \mathrm{PM} & =\frac{1}{2} O P \\ & =\frac{1}{2}(1) \\ & =\frac{1}{2} \end{aligned}$ $\mathrm{OM}=\frac{\sqrt{3}}{2} \mathrm{OP}$ $=\frac{\sqrt{3}}{2}(1)=\frac{\sqrt{3}}{2}$ Since point $P$ lies in the $2^{\text {nd }}$ quadrant, $x<0, y>0$ $\therefore x=-\mathrm{OM}=-\frac{\sqrt{3}}{2}$ and $y=\mathrm{PM}=\frac{1}{2}$ $\therefore \quad P \equiv\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ $\sin \left(-210^{\circ}\right)=y=\frac{1}{2}$ $\cos \left(-210^{\circ}\right)=x=\frac{-\sqrt{3}}{2}$ $\tan \left(-210^{\circ}\right)=\frac{y}{x}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{-\sqrt{3}}{2}\right)}=-\frac{1}{\sqrt{3}}$ $\operatorname{cosec}\left(-210^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$ $\sec \left(-210^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{-\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$ $\cot \left(-210^{\circ}\right)=\frac{x}{y}=\frac{\left(\frac{-\sqrt{3}}{2}\right)}{\left(\frac{1}{2}\right)}=-\sqrt{3}$
View full question & answer→
Question 24 Marks
Find the trigonometric functions of : -120°
Answer

Image
Angle of measure $\left(-120^{\circ}\right)$ :
Let $m \angle X O A=-120^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$
Since point $P$ lies in the $3 \mathrm{rd}$ quadrant, $x<0, y<0$
$\begin{array}{c}
\therefore \quad x=-\mathrm{OM}=\frac{-1}{2} \text { and } y=-\mathrm{PM}=\frac{-\sqrt{3}}{2} \\
\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2}\right)\\
\sin \left(-120^{\circ}\right)=y=-\frac{\sqrt{3}}{2} \\
\cos \left(-120^{\circ}\right)=x=-\frac{1}{2} \\
\tan \left(-120^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3} \\
\operatorname{cosec}\left(-120^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}} \\
\sec \left(-120^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2 \\
\left.\cot \left(-120^{\circ}\right)=\frac{x}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=\frac{1}{\sqrt{3}}\right)
\end{array}$
View full question & answer→
Question 34 Marks
Find the trigonometric functions of : 315°
Answer
Image
Angle of measure $315^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=315^{\circ}$
Its terminal arm (ray $O A)$ intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
Since point $P$ lies in the $4^{\text {th }}$ quadrant,
$x>0, y<0$
$\therefore \quad x=\mathrm{OM}=\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=-\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$\sin 315^{\circ}=y=-\frac{1}{\sqrt{2}}$
$\cos 315^{\circ}=x=\frac{1}{\sqrt{2}}$
$\tan 315^{\circ}=\frac{y}{x}=\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1$
$\operatorname{cosec} 315^{\circ}=\frac{1}{y}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$ \sec 315^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot 315^{\circ}=\frac{x}{\dot{y}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1 $
[Note: Answer given in the textbook of $\cot 315^{\circ}$ is 1. However, as per our calculation it is -1.]
View full question & answer→
Question 44 Marks
Find the trigonometric functions of : 240°
Answer
Image
Angle of measure $240^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=240^{\circ}$
Its terminal arm (ray $O A)$ intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis.
$\triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P =1$
$O M  =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2}$
$P M  =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2} $
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$ \therefore \quad x=-\mathrm{OM}=-\frac{1}{2} \text { and } y=-\mathrm{PM}=-\frac{\sqrt{3}}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2},-\frac{\sqrt{3}}{2}\right)$
$\sin 240^{\circ}=y=-\frac{\sqrt{3}}{2}$
$\cos 240^{\circ}=x=-\frac{1}{2}$
$\tan 240^{\circ}=\frac{y}{x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3}$
$\operatorname{cosec} 240^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{-\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\sec 240^{\circ}=\frac{1}{x}=\frac{1}{-\frac{1}{2}}=-2$
$\cot 240^{\circ}=\frac{x}{y}=\frac{\left(\frac{-1}{2}\right)}{\left(\frac{-\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}} $
View full question & answer→
Question 54 Marks
Find the trigonometric functions of : 225°
Answer
Image
Angle of measure $225^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=225^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg $\mathrm{PM}$ perpendicular to the $X$-axis.
$\triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$ O P =1$
$O M  =\frac{1}{\sqrt{2}} O P$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
$P M  =\frac{1}{\sqrt{2}} O P$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$ \therefore \quad x=-\mathrm{OM}=-\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=-\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$\sin 225^{\circ}=y=-\frac{1}{\sqrt{2}}$
$\cos 225^{\circ}=x=-\frac{1}{\sqrt{2}}$
$\tan 225^{\circ}=\frac{y}{x}=\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=1$
$\operatorname{cosec} 225^{\circ}=\frac{1}{y}=\frac{1}{-\frac{1}{\sqrt{2}}}=-\sqrt{2}$
$\sec 225^{\circ}=\frac{1}{x}=\frac{1}{-\frac{1}{\sqrt{2}}}=-\sqrt{2}$
$\cot 225^{\circ}=\frac{x}{y}=\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=1 $
View full question & answer→
Question 64 Marks
Find the trigonometric functions of : 120°
Answer
Image
Angle of measure $120^{\circ}$ : Let $m \angle X O A=120^{\circ}$ Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis. $\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. $O P=1$ Since point $P$ lies in the 2 nd quadrant, $x<0, y>0$
$ \therefore \quad x=-\mathrm{OM}=-\frac{1}{2} \text { and } y=\mathrm{PM}=\frac{\sqrt{3}}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2}, \frac{\sqrt{3}}{2}\right)$
$\sin 120^{\circ}=y=\frac{\sqrt{3}}{2}$
$\cos 120^{\circ}=x=-\frac{1}{2}$
$\tan 120^{\circ}=\frac{y}{x}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}$
$\operatorname{cosec} 120^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\sec 120^{\circ}=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\cot 120^{\circ}=\frac{x}{y}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}} $
[Note: Answer given in the textbook of $\tan 120^{\circ}$ is $\frac{-1}{\sqrt{3}}$ and $\cot 120^{\circ}$ is $-\sqrt{3}$.
However, as per our $-\sqrt{3}$ calculation the answer of $\tan 120^{\circ}$ is $-\sqrt{3}$ and $\cot 120^{\circ}$ is $-\frac{1}{\sqrt{3}}$
View full question & answer→
Question 74 Marks
If $\cos \theta=\frac{12}{13}, 0<\theta<\frac{\pi}{2}$ find the value of $\frac{\sin ^2 \theta-\cos ^2 \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^2 \theta}$
Answer
$\cos \theta=\frac{12}{13}$
We know that,
$ \sin ^2 \theta=1-\cos ^2 \theta$
$=1-\left(\frac{12}{13}\right)^2$
$=1-\frac{144}{169}$
$=\frac{25}{169}$
$\therefore \sin \theta= \pm \frac{5}{13} $
Since $0<\theta<\frac{\pi}{2}, \theta$ lies in the 1 st quadrant, $\therefore \sin \theta>0$
$ \therefore \quad \sin \theta=\frac{5}{13}$
$\therefore \quad \frac{\sin ^2 \theta-\cos ^2 \theta}{2 \sin \theta \cos \theta}=\frac{\left(\frac{5}{13}\right)^2-\left(\frac{12}{13}\right)^2}{2\left(\frac{5}{13}\right)\left(\frac{12}{13}\right)}$
$=\frac{\frac{25}{169}-\frac{144}{169}}{\frac{120}{169}}$
$=\frac{-\frac{119}{169}}{\frac{120}{169}}$
$=-\frac{119}{120}$
$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$
$\therefore \quad \frac{1}{\tan ^2 \theta}=\frac{1}{\left(\frac{5}{12}\right)^2}$
$=\frac{1}{\frac{25}{144}}=\frac{144}{25}$
$$
View full question & answer→
Question 84 Marks
Find the trigonometric functions of : – 240°
Answer

Image
Angle of measure (240): Let $m \angle X O A=240^{\circ}$ Its terminal arm (ray $\mathrm{OA}$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis. $\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-900$ triangle. $\begin{aligned} \mathrm{OP} & =1 \\ \mathrm{PM} & =\frac{\sqrt{3}}{2} \mathrm{OP} \\ & =\frac{\sqrt{3}}{2}(1) \\ & =\frac{\sqrt{3}}{2} \\ \mathrm{OM} & =\frac{1}{2} \mathrm{OP} \\ & =\frac{1}{2}(1)=\frac{1}{2} \end{aligned}$ Since point $P$ lies in the 2 nd quadrant, $x<0, y>0$ $\begin{array}{c} \therefore \quad x=-\mathrm{OM}=-\frac{1}{2} \text { and } y=\mathrm{PM}=\frac{\sqrt{3}}{2} \\ \therefore \quad \mathrm{P} \equiv\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \\ \sin \left(-240^{\circ}\right)=y=\frac{\sqrt{3}}{2} \\ \cos \left(-240^{\circ}\right)=x=-\frac{1}{2} \\ \tan \left(-240^{\circ}\right)=\frac{y}{x}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3} \\ \operatorname{cosec}\left(-240^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}} \\ \sec \left(-240^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2 \\ \cot \left(-240^{\circ}\right)=\frac{x}{y}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}} \end{array}$
View full question & answer→
Question 94 Marks
Find the trigonometric functions of $: – 225^\circ$
Answer
Image
Angle of measure $\left(-225^{\circ}\right)$ : Let $m \angle X O A=-225^{\circ}$ Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the X-axis. $\triangle O M P$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle. $\mathrm{OP}=1$, Since point $P$ lies in the $2$ nd quadrant, $x<0, y>0$
$ \therefore \quad x=-\mathrm{OM}=\frac{-1}{\sqrt{2}} \text { and } y=\mathrm{PM}=\frac{1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\sin \left(-225^{\circ}\right)=y=\frac{1}{\sqrt{2}}$
$\cos \left(-225^{\circ}\right)=x=-\frac{1}{\sqrt{2}}$
$\tan \left(-225^{\circ}\right)=\frac{y}{x}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1$
$\operatorname{cosec}\left(-225^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\sec \left(-225^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$\cot \left(-225^{\circ}\right)=\frac{x}{y}=\frac{1}{\frac{1}{\sqrt{2}}}=-1$
$\frac{\sqrt{2}}{} $
View full question & answer→
Question 104 Marks
Find the trigonometric functions of : – 120°
Answer

Image
Angle of measure $\left(-120^{\circ}\right)$ :
Let $m \angle X O A=-120^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$,
Since point $P$ lies in the 3rd quadrant, $x<0, y<0$
$\begin{array}{c}
\therefore \quad x=-\mathrm{OM}=\frac{-1}{2} \text { and } y=-\mathrm{PM}=\frac{-\sqrt{3}}{2} \\
\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2}\right)\\
\sin \left(-120^{\circ}\right)=y=-\frac{\sqrt{3}}{2} \\
\cos \left(-120^{\circ}\right)=x=-\frac{1}{2} \\
\tan \left(-120^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3} \\
\operatorname{cosec}\left(-120^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}} \\
\sec \left(-120^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2 \\
\cot \left(-120^{\circ}\right)=\frac{x}{y}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}
\end{array}$
View full question & answer→
Question 114 Marks
Find the trigonometric functions of $: – 45^\circ$
Answer
Image
Angle of measure $45^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=45^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$O P=1$,
Since point $P$ lies in the 4 th quadrant $x>0, y<0$
$ \therefore \quad x=\mathrm{OM}=\frac{1}{\sqrt{2}} \text { and } y=-\mathrm{PM}=\frac{-1}{\sqrt{2}}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$
$\sin \left(-45^{\circ}\right)=y=-\frac{1}{\sqrt{2}}$
$\cos \left(-45^{\circ}\right)=x=\frac{1}{\sqrt{2}}$
$\tan \left(-45^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}}\right)}=-1$
$\operatorname{cosec}\left(-45^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}=-\sqrt{2}$
$\sec \left(-45^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot \left(-45^{\circ}\right)=\frac{x}{y}=\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(-\frac{1}{\sqrt{2}}\right)}=-1 $
View full question & answer→
Question 124 Marks
Find the trigonometric functions of $: – 30^\circ$
Answer
Image
Angle of measure $30^{\circ}$
Let $m \angle X O A=-30^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60-90^{\circ}$ triangle.
$ \mathrm{Op} =1,$
$\mathrm{OP} =1,$
$\mathrm{OM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2}$
$\mathrm{PM} =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)=\frac{1}{2} $
Since point $P$ lies in the 4 th quadrant $x>0, y<0$
$ \therefore \quad x=\mathrm{OM}=\frac{\sqrt{3}}{2} \text { and } y=-\mathrm{PM}=\frac{-1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{-1}{2}\right)$
$\sin \left(-30^{\circ}\right)=y=-\frac{1}{2}$
$\cos \left(-30^{\circ}\right)=x=\frac{\sqrt{3}}{2}$
$\tan \left(-30^{\circ}\right)=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=-\frac{1}{\sqrt{3}}$
$\operatorname{cosec}\left(-30^{\circ}\right)=\frac{1}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sec \left(-30^{\circ}\right)=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot \left(-30^{\circ}\right)=\frac{x}{y}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=-\sqrt{3} $
View full question & answer→
Question 134 Marks
Find the trigonometric functions of $: 330^\circ$
Answer
Image
Angle of measure $330^{\circ}$ :
Let $m \angle X O A=330^{\circ}$
Its terminal arm (ray $\mathrm{OA})$ intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $\mathrm{X}$-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$\mathrm{OP}=1$,
Since point $P$ lies in the 4 th quadrant, $x>0, y<0$
$ \therefore \quad x=\mathrm{OM}=\frac{\sqrt{3}}{2} \text { and } y=-\mathrm{PM}=-\frac{1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{-1}{2}\right)$
$\sin 330^{\circ}=y=-\frac{1}{2}$
$\cos 330^{\circ}=x=\frac{\sqrt{3}}{2}$
$\tan 330^{\circ}=\frac{y}{x}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}$
$\operatorname{cosec} 330^{\circ}=\frac{1}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sec 330^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot 330^{\circ}=\frac{x}{y}=\frac{\sqrt{3}}{-\frac{1}{2}}=-\sqrt{3} $
View full question & answer→
Question 144 Marks
Find the trigonometric functions of $: 210^\circ$
Answer
Image
Angle of measure $210^{\circ}$ :
Let $m \angle X O A=210^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$\mathrm{OP}=1$,
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$\therefore \mathrm{x}=-\mathrm{OM}=\frac{-\sqrt{3}}{2}$ and $\mathrm{y}=-\mathrm{PM}=\frac{-1}{2}$
$\therefore P \equiv\left(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\right)$
$ \sin 210^{\circ}=y=\frac{-1}{2}$
$\cos 210^{\circ}=x=\frac{-\sqrt{3}}{2}$
$\tan 210^{\circ}=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}} $ $\tan 210^{\circ}=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}$ $ \operatorname{cosec} 210^{\circ}=\frac{1}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sec 210^{\circ}=\frac{1}{x}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\cot 210^{\circ}=\frac{x}{y}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3} $
View full question & answer→
Question 154 Marks
Find the trigonometric functions of : 150°
Answer
Image
Angle of measure $150 :$ Let $\mathrm{m} \angle \mathrm{XOA}=150^{\circ}$ Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ O P=1_r$
$\mathrm{OM}=\frac{\sqrt{3}}{2} \mathrm{OP}$
$=\frac{\sqrt{3}}{2}(1)$
$=\frac{\sqrt{3}}{2}$
$\mathrm{PM}=\frac{1}{2} \mathrm{OP}$
$=\frac{1}{2}(1)$
$=\frac{1}{2}$
$ \mathrm{OM}  =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2}$
$\mathrm{PM}  =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2} $
Since point $P$ lies in the $2$ nd quadrant, $x<0, y>0$
$ \therefore \quad x=-\mathrm{OM}=\frac{-\sqrt{3}}{2} \text { and } y=\mathrm{PM}=\frac{1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-\sqrt{3}}{2}, \frac{1}{2}\right)$
$\sin 150^{\circ}=y=\frac{1}{2}$
$\cos 150^{\circ}=x=-\frac{\sqrt{3}}{2}$
$\tan 150^{\circ}=\frac{y}{x}=\frac{\frac{1}{2}}{-\left(\frac{\sqrt{3}}{2}\right)}=-\frac{1}{\sqrt{3}}$
$\operatorname{cosec} 150^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\sec 150^{\circ}=\frac{1}{x}=\frac{1}{-\left(\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\cot 150^{\circ}=\frac{x}{y}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}}=-\sqrt{3} $
View full question & answer→
Question 164 Marks
Find the trigonometric functions of : 60°
Answer

Image
Angle of measure $60^{\circ}$ :
Let $m \angle X O A=60^{\circ}$
Its terminal arm (ray $\mathrm{OA})$ intersects the standard unit circle at $\mathrm{P}(\mathrm{x}, \mathrm{y})$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P=1$
Since point $\mathrm{P}$ lies in the $1^{\text {st }}$ quadrant, $x>0, y>0$
$\therefore \quad x=\mathrm{OM}=\frac{1}{2}$ and $y=\mathrm{PM}=\frac{\sqrt{3}}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$\sin 60^{\circ}=y=\frac{\sqrt{3}}{2}$
$\cos 60^{\circ}=x=\frac{1}{2}$
$\tan 60^{\circ}=\frac{y}{x}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2}\right)}=\sqrt{3}$
$\operatorname{cosec} 60^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\sec 60^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\cot 60^{\circ}=\frac{x}{y}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}$
View full question & answer→
Question 174 Marks
Find the trigonometric functions of : 45°
Answer
Image
Angle of measure $45^{\circ}$ :
Let $m \angle X O A=45^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle \mathrm{OMP}$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
$ \mathrm{OP}  =1,$
$\mathrm{OM}  =\frac{1}{\sqrt{2}} \mathrm{OP}$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}}$
$\mathrm{PM}  =\frac{1}{\sqrt{2}} \mathrm{OP}$
$ =\frac{1}{\sqrt{2}}(1)$
$ =\frac{1}{\sqrt{2}} $
Since point $P$ lies in the 1 st quadrant, $x>0, y>0$
$\therefore \mathrm{x}=\mathrm{OM}=\frac{1}{\sqrt{2}}$ and
$ y=P M=\frac{1}{\sqrt{2}}$
$\therefore P=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $
$\therefore \mathrm{P}=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\sin 45^{\circ}=y=\frac{1}{\sqrt{2}}$
$\cos 45^{\circ}=x=\frac{1}{\sqrt{2}}$
$\tan 45^{\circ}=\frac{y}{x}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$
$\operatorname{cosec} 45^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\sec 45^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{2}$
$\cot 45^{\circ}=\frac{x}{y}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$
View full question & answer→
Question 184 Marks
Find the trigonometric functions of $: 30^\circ$
Answer
Image
Angle of measure $30^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=30^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$ Draw seg PM perpendicular to the $X$-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$\mathrm{OP}=1$
$\mathrm{OM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2}$
$\mathrm{PM} =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2} $
Since point $P$ lies in 1st quadrant, $x>0, y>0$
$\therefore \mathrm{x}=\mathrm{OM}=\frac{\sqrt{3}}{2}$ and $\mathrm{y}=\mathrm{PM}=\frac{1}{2}$
$ \therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\sin 30^{\circ}=y=\frac{1}{2}$
$\cos 30^{\circ}=x=\frac{\sqrt{3}}{2} $
$\tan 30^{\circ}=\frac{y}{x}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$
$\operatorname{cosec} 30^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\sec 30^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot 30^{\circ}=\frac{x}{y}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
View full question & answer→
Generate Paper FreeAll Trigonometry – I topics